I understand how to get but the 3 is throwing me off a little bit, can anyone help?

The comes from partitioning the n elements into 1 set of size 3 and n - 3 sets of size 1.

Then you have to add the number of ways of partitioning the n elements into 2 sets of size 2 and n - 4 sets of size 1. This is equal to (the number of ways of choosing 4 elements from the n elements) times (number of ways of choosing 2 elements from 4)/2:

.

You divide by 2 because otherwise you've double counted .... Think about it.

I understand how to get but the 3 is throwing me off a little bit, can anyone help?

First, these must be Stirling Numbers of the second kind.S(n,k) is the number of ways to partition a set of n individuals into k non-empty cells.
Therefore, in this problem we must have .
So if then how many ways can one partition the set in two non-empty subsets?
It is clear that we can have one cell with three elements and one with one element: ways.

First, these must be Stirling Numbers of the second kind.S(n,k) is the number of ways to partition a set of n individuals into k non-empty cells.
Therefore, in this problem we must have .
So if then how many ways can one partition the set in two non-empty subsets?
It is clear that we can have one cell with three elements and one with one element: ways.